Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline int chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
inline int chkmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const int maxn = 100000 + 15;
const int maxk = 10;
const int logn = 19;
int n, m, q, lev[maxn], par[maxn][logn], mark[maxn], tick;
vector<int> g[maxn];
vector<pair<int, int> > f[maxn][logn], a[maxn];
vector<pair<int, int> > join(vector<pair<int, int> > a,
                             vector<pair<int, int> > b) {
  tick++;
  vector<pair<int, int> > ret;
  int i = 0, j = 0;
  while (i < int((a).size()) && j < int((b).size())) {
    if (mark[a[i].first] == tick) {
      i++;
      continue;
    }
    if (mark[b[j].first] == tick) {
      j++;
      continue;
    }
    if (a[i] <= b[j]) {
      if (int((ret).size()) < maxk) ret.push_back(a[i]);
      mark[a[i].first] = tick;
      i++;
    } else {
      if (int((ret).size()) < maxk) ret.push_back(b[j]);
      mark[b[j].first] = tick;
      j++;
    }
  }
  while (i < int((a).size())) {
    if (mark[a[i].first] == tick) {
      i++;
      continue;
    }
    if (int((ret).size()) < maxk) ret.push_back(a[i]);
    mark[a[i].first] = tick;
    i++;
  }
  while (j < int((b).size())) {
    if (mark[b[j].first] == tick) {
      j++;
      continue;
    }
    if (int((ret).size()) < maxk) ret.push_back(b[j]);
    mark[b[j].first] = tick;
    j++;
  }
  return ret;
}
void dfs(int u, int p, int d) {
  lev[u] = d;
  for (int i = (0), _b = (int((g[u]).size()) - 1); i <= _b; ++i) {
    int v = g[u][i];
    if (v == p) continue;
    par[v][0] = u;
    f[v][0] = join(a[u], a[v]);
    dfs(v, u, d + 1);
  }
}
void init() {
  memset(par, -1, sizeof(par));
  tick = 0;
  dfs(1, 1, 1);
  for (int j = (1), _b = (logn - 1); j <= _b; ++j)
    for (int i = (1), _b = (n); i <= _b; ++i)
      if (~par[i][j - 1]) {
        par[i][j] = par[par[i][j - 1]][j - 1];
        f[i][j] = join(f[i][j - 1], f[par[i][j - 1]][j - 1]);
      }
}
vector<pair<int, int> > query(int u, int v) {
  if (lev[u] < lev[v]) return query(v, u);
  vector<pair<int, int> > ret;
  ret = join(ret, a[u]);
  ret = join(ret, a[v]);
  for (int i = (logn - 1), _b = (0); i >= _b; --i)
    if (lev[par[u][i]] >= lev[v]) {
      ret = join(ret, f[u][i]);
      u = par[u][i];
    }
  if (u == v) {
    return ret;
  }
  for (int i = (logn - 1), _b = (0); i >= _b; --i)
    if (par[u][i] != par[v][i]) {
      ret = join(ret, f[u][i]);
      ret = join(ret, f[v][i]);
      u = par[u][i];
      v = par[v][i];
    }
  ret = join(ret, f[u][0]);
  ret = join(ret, f[v][0]);
  return ret;
}
void solve(int u, int v, int k) {
  vector<pair<int, int> > ans = query(u, v);
  printf("%d ", min(int((ans).size()), k));
  for (int i = (0), _b = (min(int((ans).size()), k) - 1); i <= _b; ++i)
    printf("%d ", ans[i].first);
  putchar('\n');
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = (1), _b = (n - 1); i <= _b; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = (1), _b = (m); i <= _b; ++i) {
    int c;
    scanf("%d", &c);
    if (int((a[c]).size()) < 10) a[c].push_back(make_pair(i, c));
  }
  init();
  while (q--) {
    int u, v, k;
    scanf("%d %d %d", &u, &v, &k);
    solve(u, v, k);
  }
  return 0;
}